- #1
Dustinsfl
- 2,281
- 5
Suppose $\lim\limits_{n\to\infty}a_n = a$. Prove using the delta-epsilon definition that $\lim\limits_{n\to\infty}a_n^2 = a^2$.
Let $\epsilon > 0$ be given. Let $N\in\mathbb{Z}$. Then for $n > N$, $|a_n - a| < \epsilon$.
Then
$$
|a_n^2 - a^2| = |(a_n - a)(a_n + a)| < \epsilon |a_n + a|
$$
I am stuck here.
Let $\epsilon > 0$ be given. Let $N\in\mathbb{Z}$. Then for $n > N$, $|a_n - a| < \epsilon$.
Then
$$
|a_n^2 - a^2| = |(a_n - a)(a_n + a)| < \epsilon |a_n + a|
$$
I am stuck here.